- Thread starter
- #1

- Thread starter Shobhit
- Start date

- Thread starter
- #1

- Mar 22, 2013

- 573

Very nice, I like this one. The proof follows immediately from basic genus theory, and I am not going to post it down here (spoiler purpose).

For beginners, I would give a hint :

A prime of the particular form is of the form 1, 3, 7 or 9 modulo 20

Balarka

.

- Feb 13, 2012

- 1,704

If $p$ is an odd prime expressible as

$$p=x^2+5y^2$$

where $x,y$ are integers, then prove that $p \equiv 1 \text{ or }9\text{ (mod 20)}$.

$\displaystyle p = 4\ n^{2} + 20\ m^{2} + 20\ m + 5\ (1)$

... and then...

$\displaystyle p \equiv 4\ n^{2} + 5\ \text{mod}\ 20\ (2)$

Observing (2) it is evident that $\displaystyle p \equiv 1\ \text {mod}\ 20$ or $\displaystyle p \equiv 9\ \text {mod}\ 20$. The same result You obtaining supposing x odd and y even...

Kind regards

$\chi$ $\sigma$

Last edited:

- Mar 22, 2013

- 573

How does (2) follows from (1)?

- Feb 13, 2012

- 1,704

I made a mistake writing $\displaystyle p \equiv 4\ n^{2}\ \text{mod}\ 20$ instead of $\displaystyle p \equiv 4\ n^{2} + 5\ \text{mod}\ 20$, so that I corrected it... very sorry! ...How does (2) follows from (1)?

Kind regards

$\chi$ $\sigma$

- Mar 22, 2013

- 573

Ah, I see, I have interpreted the question wrong. I interpreted is as : p is an odd prime expressible as x^2 + 5y^2 if and only if p is 1 or 9 modulo 20.

But perhaps OP must have intended exactly this? Can Shobhit clarify whether he did a typo or not?

- Thread starter
- #7

Balarka, the question is correct. You had interpreted it correctly.Ah, I see, I have interpreted the question wrong. I interpreted is as : p is an odd prime expressible as x^2 + 5y^2 if and only if p is 1 or 9 modulo 20.

But perhaps OP must have intended exactly this? Can Shobhit clarify whether he did a typo or not?

- Mar 22, 2013

- 573

I see, thank you for clarifying.Shobhit said:Balarka, the question is correct. You had interpreted it correctly.

One can carry on with my hints on the #2 then.

Last edited: